What is the missing constant term in the perfect square that starts with $x^2-8x$ ?
Explanation: Let $b$ be the missing constant term. Let's assume $x^2-8x+b$ is factored as the perfect square $(x+a)^2$. $\begin{aligned} (x+a)^2&=x^2+{2a}x+{a^2} \\\\ &=x^2{-8}x+ b \end{aligned}$ For the expressions to be the same, ${2a}$ must be equal to ${-8}$, and ${a^2}$ must be equal to $ b$. From ${2a=-8}$ we know that $a=-4$. Now, from ${a^2=b}$ we know that $b=(-4)^2=16$. Indeed, $x^2-8x+16$ is factored as $(x-4)^2$. In conclusion, the missing constant term in the perfect square that starts with $x^2-8x$ is $16$